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Chapter 11: Problem 24

Determine whether the sequence converges or diverges. If it converges, find the limit. $$ a_{n}=\sqrt{\frac{n+1}{9 n+1}} $$

Short Answer

Expert verified
The sequence converges to \( \frac{1}{3} \).

Step by step solution

01

Understand Sequence Convergence

A sequence converges if its terms approach a single, finite limit as \( n \to \infty \). First, we need to analyze the behavior of the sequence \( a_n = \sqrt{\frac{n+1}{9n+1}} \) as \( n \) becomes very large.
02

Simplify the Expression

Begin by rewriting the inside of the square root to highlight dominant terms. Consider \( \frac{n+1}{9n+1} = \frac{n(1 + \frac{1}{n})}{9n(1 + \frac{1}{9n})} = \frac{1 + \frac{1}{n}}{9 + \frac{1}{9n}} \). As \( n \to \infty \), the fractional parts \( \frac{1}{n} \) and \( \frac{1}{9n} \) approach 0.
03

Approximate Limit

As \( n \to \infty \), the dominant terms in the numerator and denominator simplify to \( \frac{1}{9} \). Therefore, \( \lim_{{n \to \infty}} \frac{1 + \frac{1}{n}}{9 + \frac{1}{9n}} = \frac{1}{9} \). Take the square root to find \( \lim_{{n \to \infty}} a_n = \sqrt{\frac{1}{9}} = \frac{1}{3} \).
04

Conclusion

We determined that as \( n \to \infty \), \( a_n \) approaches \( \frac{1}{3} \). Thus, the sequence converges to \( \frac{1}{3} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limit of a Sequence
In mathematics, understanding the limit of a sequence is crucial for determining its long-term behavior. A sequence converges when its terms get closer and closer to a specific number as the term number "n" increases indefinitely. This specific number is known as the limit of the sequence.

When analyzing the sequence \( a_n = \sqrt{\frac{n+1}{9n+1}} \), we're interested in what happens to \( a_n \) as \( n \to \infty \):
  • We observe that evaluating limits often involves simplifying terms to see which parts dominate as \( n \) grows larger.
  • For this sequence, as \( n \to \infty \), the effect of smaller terms like \( \frac{1}{n} \) becomes negligible compared to dominant terms.
By simplifying to highlight these terms, we were able to see the behavior clearly, and hence, the sequence converged to the limit \( \frac{1}{3} \).
Dominant Terms
When evaluating sequences, especially in the limit, recognizing dominant terms is a powerful tool. Dominant terms are those that have the most significant impact on the expression as \( n \to \infty \). In our example, the sequence \( a_n = \sqrt{\frac{n+1}{9n+1}} \) requires us to simplify to find dominant terms.
  • Initially, we rewrite \( \frac{n+1}{9n+1} \) to emphasize what controls the value as \( n \to \infty \).
  • This involves factoring out \( n \) from both the numerator and denominator, revealing \( \frac{1 + \frac{1}{n}}{9 + \frac{1}{9n}} \).
As \( n \to \infty \), the smaller terms \( \frac{1}{n} \) and \( \frac{1}{9n} \) diminish to zero, leaving us primarily with the dominant terms \( 1 \) and \( 9 \). This leads us to realize that the critical behavior of the fraction is determined by these dominant terms, simplifying our understanding of the sequence's behavior.
Sequence Behavior as n Approaches Infinity
Evaluating how a sequence behaves as "n approaches infinity" allows us to see its ultimate trend. For \( a_n = \sqrt{\frac{n+1}{9n+1}} \), the behavior can be predicted once dominant terms are identified. As \( n \to \infty \), we concentrate on the terms that don't diminish:
  • The terms \( \frac{1}{n} \) and \( \frac{1}{9n} \) become less significant or vanish as "\( n \to \infty \)".
  • Hence, we focus on the core ratio of \( 1 \) to \( 9 \), which simplifies to \( \frac{1}{9} \).
  • Once we take the square root of this simplified form, we're left with \( \frac{1}{3} \), representing the stable value or convergence point of \( a_n \).
In conclusion, the sequence steadily approaches \( \frac{1}{3} \) as \( n \) becomes very large. Recognizing this behavior is essential for predicting and understanding the convergence of sequences.

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